\(\int \frac {\sin ^2(c+d x)}{a+b \tan (c+d x)} \, dx\) [54]

Optimal result

Integrand size = 21, antiderivative size = 94 \[ \int \frac {\sin ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {a \left (a^2-b^2\right ) x}{2 \left (a^2+b^2\right )^2}+\frac {a^2 b \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d} \]

[Out]

1/2*a*(a^2-b^2)*x/(a^2+b^2)^2+a^2*b*ln(a*cos(d*x+c)+b*sin(d*x+c))/(a^2+b^2)^2/d-1/2*cos(d*x+c)^2*(b+a*tan(d*x+
c))/(a^2+b^2)/d

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3597, 1661, 815, 649, 209, 266} \[ \int \frac {\sin ^2(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {\cos ^2(c+d x) (a \tan (c+d x)+b)}{2 d \left (a^2+b^2\right )}+\frac {a^2 b \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac {a x \left (a^2-b^2\right )}{2 \left (a^2+b^2\right )^2} \]

[In]

Int[Sin[c + d*x]^2/(a + b*Tan[c + d*x]),x]

[Out]

(a*(a^2 - b^2)*x)/(2*(a^2 + b^2)^2) + (a^2*b*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)^2*d) - (Cos[c
+ d*x]^2*(b + a*Tan[c + d*x]))/(2*(a^2 + b^2)*d)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1661

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[(a*g - c*f*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 3597

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b/f, Subst[Int
[x^m*((a + x)^n/(b^2 + x^2)^(m/2 + 1)), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/
2]

Rubi steps \begin{align*} \text {integral}& = \frac {b \text {Subst}\left (\int \frac {x^2}{(a+x) \left (b^2+x^2\right )^2} \, dx,x,b \tan (c+d x)\right )}{d} \\ & = -\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d}-\frac {\text {Subst}\left (\int \frac {-\frac {a^2 b^2}{a^2+b^2}+\frac {a b^2 x}{a^2+b^2}}{(a+x) \left (b^2+x^2\right )} \, dx,x,b \tan (c+d x)\right )}{2 b d} \\ & = -\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d}-\frac {\text {Subst}\left (\int \left (-\frac {2 a^2 b^2}{\left (a^2+b^2\right )^2 (a+x)}-\frac {a b^2 \left (a^2-b^2-2 a x\right )}{\left (a^2+b^2\right )^2 \left (b^2+x^2\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{2 b d} \\ & = \frac {a^2 b \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d}+\frac {(a b) \text {Subst}\left (\int \frac {a^2-b^2-2 a x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d} \\ & = \frac {a^2 b \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d}-\frac {\left (a^2 b\right ) \text {Subst}\left (\int \frac {x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{\left (a^2+b^2\right )^2 d}+\frac {\left (a b \left (a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d} \\ & = \frac {a \left (a^2-b^2\right ) x}{2 \left (a^2+b^2\right )^2}+\frac {a^2 b \log (\cos (c+d x))}{\left (a^2+b^2\right )^2 d}+\frac {a^2 b \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^2 d}-\frac {\cos ^2(c+d x) (b+a \tan (c+d x))}{2 \left (a^2+b^2\right ) d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.81 \[ \int \frac {\sin ^2(c+d x)}{a+b \tan (c+d x)} \, dx=-\frac {2 a b \left (a^2+b^2\right ) \arctan (\tan (c+d x))+2 b^2 \left (a^2+b^2\right ) \cos ^2(c+d x)+a \left (2 a \left (\left (b^2+a \sqrt {-b^2}\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )-2 b^2 \log (a+b \tan (c+d x))+\left (b^2-a \sqrt {-b^2}\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )\right )+b \left (a^2+b^2\right ) \sin (2 (c+d x))\right )}{4 b \left (a^2+b^2\right )^2 d} \]

[In]

Integrate[Sin[c + d*x]^2/(a + b*Tan[c + d*x]),x]

[Out]

-1/4*(2*a*b*(a^2 + b^2)*ArcTan[Tan[c + d*x]] + 2*b^2*(a^2 + b^2)*Cos[c + d*x]^2 + a*(2*a*((b^2 + a*Sqrt[-b^2])
*Log[Sqrt[-b^2] - b*Tan[c + d*x]] - 2*b^2*Log[a + b*Tan[c + d*x]] + (b^2 - a*Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Ta
n[c + d*x]]) + b*(a^2 + b^2)*Sin[2*(c + d*x)]))/(b*(a^2 + b^2)^2*d)

Maple [A] (verified)

Time = 1.15 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.30

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (-\frac {1}{2} a^{3}-\frac {1}{2} a \,b^{2}\right ) \tan \left (d x +c \right )-\frac {a^{2} b}{2}-\frac {b^{3}}{2}}{1+\tan ^{2}\left (d x +c \right )}+\frac {a \left (-a b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+\left (a^{2}-b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )\right )}{2}}{\left (a^{2}+b^{2}\right )^{2}}+\frac {a^{2} b \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(122\)
default	\(\frac {\frac {\frac {\left (-\frac {1}{2} a^{3}-\frac {1}{2} a \,b^{2}\right ) \tan \left (d x +c \right )-\frac {a^{2} b}{2}-\frac {b^{3}}{2}}{1+\tan ^{2}\left (d x +c \right )}+\frac {a \left (-a b \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+\left (a^{2}-b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )\right )}{2}}{\left (a^{2}+b^{2}\right )^{2}}+\frac {a^{2} b \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(122\)
risch	\(-\frac {a x}{2 \left (2 i a b -a^{2}+b^{2}\right )}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 \left (-i b +a \right ) d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 \left (i b +a \right ) d}-\frac {2 i a^{2} b x}{a^{4}+2 a^{2} b^{2}+b^{4}}-\frac {2 i a^{2} b c}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {a^{2} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{d \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}\)	\(175\)

[In]

int(sin(d*x+c)^2/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/(a^2+b^2)^2*(((-1/2*a^3-1/2*a*b^2)*tan(d*x+c)-1/2*a^2*b-1/2*b^3)/(1+tan(d*x+c)^2)+1/2*a*(-a*b*ln(1+tan(
d*x+c)^2)+(a^2-b^2)*arctan(tan(d*x+c))))+a^2*b/(a^2+b^2)^2*ln(a+b*tan(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.30 \[ \int \frac {\sin ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {a^{2} b \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) + {\left (a^{3} - a b^{2}\right )} d x - {\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} \]

[In]

integrate(sin(d*x+c)^2/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(a^2*b*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) + (a^3 - a*b^2)*d*x - (a^2*
b + b^3)*cos(d*x + c)^2 - (a^3 + a*b^2)*cos(d*x + c)*sin(d*x + c))/((a^4 + 2*a^2*b^2 + b^4)*d)

Sympy [F]

\[ \int \frac {\sin ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\int \frac {\sin ^{2}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \]

[In]

integrate(sin(d*x+c)**2/(a+b*tan(d*x+c)),x)

[Out]

Integral(sin(c + d*x)**2/(a + b*tan(c + d*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.53 \[ \int \frac {\sin ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {2 \, a^{2} b \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {a^{2} b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a^{3} - a b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {a \tan \left (d x + c\right ) + b}{{\left (a^{2} + b^{2}\right )} \tan \left (d x + c\right )^{2} + a^{2} + b^{2}}}{2 \, d} \]

[In]

integrate(sin(d*x+c)^2/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/2*(2*a^2*b*log(b*tan(d*x + c) + a)/(a^4 + 2*a^2*b^2 + b^4) - a^2*b*log(tan(d*x + c)^2 + 1)/(a^4 + 2*a^2*b^2
+ b^4) + (a^3 - a*b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) - (a*tan(d*x + c) + b)/((a^2 + b^2)*tan(d*x + c)^2 +
a^2 + b^2))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (90) = 180\).

Time = 0.39 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.96 \[ \int \frac {\sin ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {2 \, a^{2} b^{2} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} - \frac {a^{2} b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {{\left (a^{3} - a b^{2}\right )} {\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {a^{2} b \tan \left (d x + c\right )^{2} - a^{3} \tan \left (d x + c\right ) - a b^{2} \tan \left (d x + c\right ) - b^{3}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (d x + c\right )^{2} + 1\right )}}}{2 \, d} \]

[In]

integrate(sin(d*x+c)^2/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

1/2*(2*a^2*b^2*log(abs(b*tan(d*x + c) + a))/(a^4*b + 2*a^2*b^3 + b^5) - a^2*b*log(tan(d*x + c)^2 + 1)/(a^4 + 2
*a^2*b^2 + b^4) + (a^3 - a*b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) + (a^2*b*tan(d*x + c)^2 - a^3*tan(d*x + c) -
a*b^2*tan(d*x + c) - b^3)/((a^4 + 2*a^2*b^2 + b^4)*(tan(d*x + c)^2 + 1)))/d

Mupad [B] (verification not implemented)

Time = 5.01 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.56 \[ \int \frac {\sin ^2(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {a^2\,b\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d\,{\left (a^2+b^2\right )}^2}-\frac {a\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{4\,d\,\left (-a^2\,1{}\mathrm {i}+2\,a\,b+b^2\,1{}\mathrm {i}\right )}-\frac {{\cos \left (c+d\,x\right )}^2\,\left (\frac {b}{2\,\left (a^2+b^2\right )}+\frac {a\,\mathrm {tan}\left (c+d\,x\right )}{2\,\left (a^2+b^2\right )}\right )}{d}-\frac {a\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,d\,\left (-a^2+a\,b\,2{}\mathrm {i}+b^2\right )} \]

[In]

int(sin(c + d*x)^2/(a + b*tan(c + d*x)),x)

[Out]

(a^2*b*log(a + b*tan(c + d*x)))/(d*(a^2 + b^2)^2) - (a*log(tan(c + d*x) + 1i)*1i)/(4*d*(a*b*2i - a^2 + b^2)) -
(a*log(tan(c + d*x) - 1i))/(4*d*(2*a*b - a^2*1i + b^2*1i)) - (cos(c + d*x)^2*(b/(2*(a^2 + b^2)) + (a*tan(c +
d*x))/(2*(a^2 + b^2))))/d